Problem: What are the first three non-zero terms of the Maclaurin series for $\dfrac{\sin(x)}{x}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ x-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}$ (Choice B) B $\frac{1}{x}+\frac{x}{2!}+\frac{{{x}^{3}}}{4!}$ (Choice C) C $1+\frac{{{x}^{2}}}{3!}+\frac{{{x}^{4}}}{5!}$ (Choice D) D $\frac{1}{x}-\frac{x}{2!}+\frac{{{x}^{3}}}{4!}$ (Choice E) E $ 1-\frac{{{x}^{2}}}{3!}+\frac{{{x}^{4}}}{5!}$
Start with the Maclaurin series for $\sin x$. $\sin x=x-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}-...+{{\left( -1 \right)}^{n}}\frac{{{x}^{2n+1}}}{\left( 2n+1 \right)!}+...$ Divide by $x$. $\frac{\sin x}{x}=1-\frac{{{x}^{2}}}{3!}+\frac{{{x}^{4}}}{5!}-...+{{\left( -1 \right)}^{n}}\frac{{{x}^{2n}}}{\left( 2n+1 \right)!}+...$ The first three terms are $1~-~\dfrac{x^2}{3!}~+~\dfrac{x^4}{5!}$.